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P=3 /1.22 +1/22.32+...+4033/20162.20172
P=1/1 -1/22 +1/22 -1/52 +...+1/20162 - 1/20172
P=1-1/20172 <1
vậy p<1
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Ta có nhận xét sau
Với n là số nguyên dương
\(\frac{2n+1}{n^2.\left(n+1\right)^2}=\frac{n^2+2n+1-n^2}{n^2\left(n+1\right)^2}=\frac{\left(n+1\right)^2}{n^2\left(n+1\right)^2}-\frac{n^2}{n^2\left(n+1\right)^2}=\frac{1}{n^2}-\frac{1}{\left(n+1\right)^2}\)
Áp dụng nhận xét trên ,ta có
\(P=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+....+\frac{1}{2016^2}-\frac{1}{2017^2}=1-\frac{1}{2017^2}>1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(P=\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(2.3\right)^2}+\frac{7}{\left(3.4\right)^2}+.....+\frac{4033}{\left(2016.2017\right)^2}\)
\(\frac{2^2-1^2}{1^2.2^2}+\frac{3^2-2^2}{2^2.3^2}+\frac{4^2-3^2}{3^2.4^2}+.....+\frac{2017^2-2016^2}{2016^2.2017^2}\)
\(=1-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+....+\frac{1}{2016^2}-\frac{1}{2017^2}\)
\(=1-\frac{1}{2017^2}< 11\) (đpcm)
Bài này trong đề thi học kì 2 môn Toán lớp 6 trường Amsterdam năm 2016-2017 này. Mình 10 luôn hehe
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\(A=\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2016.2017}\right):2\)
\(=\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2016}-\frac{1}{2017}\right):2\)
\(=\left(1-\frac{1}{2017}\right):2\)\(< \)\(\frac{1}{2}\) (Do 1 - 1/2017 < 1)
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11.2+12.3+13.4+14.5+...+12015.2016+12016.2017
=1−12+12−13+13−14+14−15+...+12015−12016+12016−12017
=1−12017=20162017
![](https://rs.olm.vn/images/avt/0.png?1311)
A = \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2016.2017}\)
A = \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2016}-\frac{1}{2017}\)
A = \(1-\left(\frac{1}{2}-\frac{1}{2}\right)-\left(\frac{1}{3}-\frac{1}{3}\right)-...-\left(\frac{1}{2016}-\frac{1}{2016}\right)-\frac{1}{2017}\)
A = \(1-0-0-0...-0-\frac{1}{2017}\)
A = \(1-\frac{1}{2017}< 1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
A=1.2+2.3+3.4+4.5+5.6+...+2016.2017
=> 3A = 1.2.3+2.3.3+3.4.3+4.5.3+5.6.3+.......+2016.2017.3
=> 3A = 1.2.3 + 2.3.(4-1) + 3.4.(5-2) + 4.5.(6-3) + .......+ 2016.2017.(2018-2015)
=> 3A = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 +..........+ 2016.2017.2018 - 2015.2016.2017
=> 3A = 2016.2017.2018
=> A = 2016.2017.2018 : 3
\(A=\dfrac{3}{\left(1.2\right)^2}+\dfrac{5}{\left(2.3\right)^2}+...+\dfrac{4033}{\left(2016.2017\right)^2}\)
\(=\dfrac{3}{1.2^2}+\dfrac{5}{2^2.3^2}+...+\dfrac{4033}{2016^2.2017^2}\)
\(=\dfrac{1}{1}-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+...+\dfrac{1}{2016^2}-\dfrac{1}{2017^2}\)
\(=1-\dfrac{1}{2017^2}< 1\)
\(\Rightarrow A< 1\left(đpcm\right)\)
Vậy...